Question

A certain substance X decomposes. It is found that 50% of X remains after 100 minutes. How much X remains after 200 minutes if the reaction order with respect to X is a) zero order b) first order c) second order

Answer 1

Zero Order: t(1/2) = [A]o/(2*k), so k = [A]o/(2 * t(1/2)), where t(1/2) = 100 min. and [A]o=100% So incorporate condition. A= - .0005 x 200 mins+1 of introductory fixation approaches zero percent. 
Initially Order: Half-life condition k=.00693 remaining with 100 mins half life 
Coordinate condition lnx= - .00693x200mins + ln 1 squares with lnx= - 1.386. Take to the e to counteract ln so x=.25 or 25% 
Second Order Initial Concentration of 1, half-existence of 100 mins. K= .01. Coordinate condition 1/x = .01 x 200 +1. X= 33.3%